Definition and ROC
Bilateral Z-Transform
X(z) = \sum_{n=-\infty}^{\infty} x[n] z^{-n}
| Symbol | Description | Unit |
|---|---|---|
| X(z) | Z-transform of sequence | — |
| x[n] | Discrete-time sequence | — |
| z | Complex variable | — |
Worked example
Find the Z-transform of x[n] = {1, 2, 3} (causal, n = 0, 1, 2).
Given: x[0]=1, x[1]=2, x[2]=3
- Apply X(z) = sum of x[n] z^{-n}
- X(z) = x[0]z^0 + x[1]z^{-1} + x[2]z^{-2}
- X(z) = 1 + 2z^{-1} + 3z^{-2}
- ROC: entire z-plane except z = 0
Answer: X(z) = 1 + 2z^{-1} + 3z^{-2}, ROC: z ≠ 0
Unilateral Z-Transform
X(z) = \sum_{n=0}^{\infty} x[n] z^{-n}
| Symbol | Description | Unit |
|---|---|---|
| X(z) | One-sided Z-transform | — |
| x[n] | Causal sequence | — |
Worked example
Find the unilateral Z-transform of x[n] = (0.5)^n u[n].
Given: a = 0.5
- X(z) = sum_{n=0}^{inf} (0.5)^n z^{-n}
- = sum_{n=0}^{inf} (0.5/z)^n
- This is a geometric series with ratio r = 0.5z^{-1}
- Converges when |0.5z^{-1}| < 1, i.e., |z| > 0.5
- X(z) = 1/(1 - 0.5z^{-1}) = z/(z - 0.5)
Answer: X(z) = z/(z - 0.5), ROC: |z| > 0.5
ROC for Causal Sequence
\text{ROC: } |z| > r_{\max}
| Symbol | Description | Unit |
|---|---|---|
| r_max | Magnitude of outermost pole | — |
Worked example
A causal sequence has poles at z = 0.8 and z = -0.5. Determine the ROC.
Given: poles at z = 0.8 and z = -0.5
- For a causal (right-sided) sequence, ROC is exterior to outermost pole
- Pole magnitudes: |0.8| = 0.8, |-0.5| = 0.5
- Outermost pole has magnitude r_max = 0.8
- ROC: |z| > 0.8
Answer: ROC: |z| > 0.8
Standard Z-Transform Pairs
Unit Impulse
\delta[n] \xleftrightarrow{Z} 1, \quad \text{ROC: all } z
| Symbol | Description | Unit |
|---|---|---|
| δ[n] | Unit impulse sequence | — |
Worked example
Verify Z{δ[n-2]}.
Given: Delayed impulse by 2 samples
- Z{δ[n-2]} = sum_{n=-inf}^{inf} δ[n-2] z^{-n}
- Only n=2 contributes: δ[0] = 1
- = 1 * z^{-2} = z^{-2}
Answer: Z{δ[n-2]} = z^{-2}
Exponential Sequence
a^n u[n] \xleftrightarrow{Z} \frac{z}{z-a}, \quad |z| > |a|
| Symbol | Description | Unit |
|---|---|---|
| a | Base of exponential | — |
| u[n] | Unit step sequence | — |
Worked example
Find X(z) for x[n] = (0.25)^n u[n].
Given: a = 0.25
- Use pair: a^n u[n] <-> z/(z-a)
- Substitute a = 0.25
- X(z) = z/(z - 0.25)
- ROC: |z| > 0.25
Answer: X(z) = z/(z - 0.25), ROC: |z| > 0.25
Unit Step Sequence
u[n] \xleftrightarrow{Z} \frac{z}{z-1}, \quad |z| > 1
| Symbol | Description | Unit |
|---|---|---|
| u[n] | Unit step, u[n]=1 for n≥0 | — |
Worked example
Find the Z-transform of x[n] = 3u[n].
Given: Amplitude = 3
- By linearity: Z{3u[n]} = 3 * Z{u[n]}
- Z{u[n]} = z/(z-1)
- X(z) = 3z/(z-1)
- ROC: |z| > 1
Answer: X(z) = 3z/(z-1), ROC: |z| > 1
Ramp Sequence
n \cdot u[n] \xleftrightarrow{Z} \frac{z}{(z-1)^2}, \quad |z| > 1
| Symbol | Description | Unit |
|---|---|---|
| n·u[n] | Discrete ramp sequence | — |
Worked example
Find Z-transform of x[n] = 2n·u[n].
Given: Coefficient = 2
- Z{2n u[n]} = 2 Z{n u[n]}
- Z{n u[n]} = z/(z-1)^2
- X(z) = 2z/(z-1)^2
- ROC: |z| > 1
Answer: X(z) = 2z/(z-1)^2, ROC: |z| > 1
Cosine Sequence
\cos(\omega_0 n)u[n] \xleftrightarrow{Z} \frac{z(z-\cos\omega_0)}{z^2 - 2z\cos\omega_0 + 1}
| Symbol | Description | Unit |
|---|---|---|
| ω₀ | Digital frequency | rad/sample |
Worked example
Find Z-transform of cos(π/4 · n)u[n].
Given: ω₀ = π/4, cos(π/4) = 0.707
- Numerator: z(z - cos(π/4)) = z(z - 0.707) = z^2 - 0.707z
- Denominator: z^2 - 2(0.707)z + 1 = z^2 - 1.414z + 1
- X(z) = (z^2 - 0.707z)/(z^2 - 1.414z + 1)
Answer: X(z) = (z^2 - 0.707z)/(z^2 - 1.414z + 1), ROC: |z| > 1
Z-Transform Properties
Time Shifting
x[n-k] \xleftrightarrow{Z} z^{-k} X(z)
| Symbol | Description | Unit |
|---|---|---|
| k | Number of samples of delay | samples |
| z^{-k} | Delay operator | — |
Worked example
If X(z) = z/(z-0.5), find Z-transform of x[n-3].
Given: k = 3, X(z) = z/(z-0.5)
- Apply time-shift property: Z{x[n-3]} = z^{-3} X(z)
- = z^{-3} * z/(z-0.5)
- = z^{-2}/(z-0.5)
- = 1/(z^2(z-0.5))
Answer: Z{x[n-3]} = z^{-2}/(z - 0.5)
Convolution in Time
x_1[n] * x_2[n] \xleftrightarrow{Z} X_1(z) \cdot X_2(z)
| Symbol | Description | Unit |
|---|---|---|
| * | Convolution operator | — |
Worked example
Find y[n] = x1[n]*x2[n] where X1(z)=z/(z-0.5) and X2(z)=z/(z-0.25).
Given: X1(z) = z/(z-0.5), X2(z) = z/(z-0.25)
- Y(z) = X1(z) X2(z) = z^2/((z-0.5)(z-0.25))
- Partial fractions: Y(z)/z = z/((z-0.5)(z-0.25))
- A/(z-0.5) + B/(z-0.25)
- A = 0.5/(0.5-0.25) = 2, B = 0.25/(0.25-0.5) = -1
- Y(z) = 2z/(z-0.5) - z/(z-0.25)
- y[n] = [2(0.5)^n - (0.25)^n]u[n]
Answer: y[n] = [2(0.5)^n - (0.25)^n]u[n]
Differentiation in z-Domain
n \cdot x[n] \xleftrightarrow{Z} -z \frac{dX(z)}{dz}
| Symbol | Description | Unit |
|---|---|---|
| dX(z)/dz | Derivative of X(z) w.r.t. z | — |
Worked example
Find Z-transform of n·(0.5)^n·u[n] using the differentiation property.
Given: a = 0.5, Z{(0.5)^n u[n]} = z/(z-0.5)
- Let X(z) = z/(z-0.5)
- dX/dz = [(z-0.5)(1) - z(1)]/(z-0.5)^2 = -0.5/(z-0.5)^2
- Z{n(0.5)^n u[n]} = -z * (-0.5/(z-0.5)^2)
- = 0.5z/(z-0.5)^2
Answer: Z{n(0.5)^n u[n]} = 0.5z/(z-0.5)^2
Initial Value Theorem
x[0] = \lim_{z \to \infty} X(z)
| Symbol | Description | Unit |
|---|---|---|
| x[0] | Initial value of sequence | — |
Worked example
Given X(z) = (3z^2 - z)/(z^2 - 0.5z + 0.25), find x[0].
Given: X(z) = (3z^2 - z)/(z^2 - 0.5z + 0.25)
- Apply IVT: x[0] = lim_{z->inf} X(z)
- Divide numerator and denominator by z^2
- = lim_{z->inf} (3 - 1/z)/(1 - 0.5/z + 0.25/z^2)
- = (3 - 0)/(1 - 0 + 0) = 3
Answer: x[0] = 3
Final Value Theorem
x[\infty] = \lim_{z \to 1} (z-1) X(z)
| Symbol | Description | Unit |
|---|---|---|
| x[∞] | Steady-state value of sequence | — |
Worked example
Find the final value of x[n] if X(z) = z/(z-1)(z-0.5).
Given: X(z) = z/((z-1)(z-0.5))
- Check FVT applicability: poles of (z-1)X(z) must be inside unit circle
- (z-1)X(z) = z/((z-0.5)) — pole at z=0.5, |0.5|<1, FVT valid
- x[inf] = lim_{z->1} (z-1) * z/((z-1)(z-0.5))
- = lim_{z->1} z/(z-0.5) = 1/(1-0.5) = 2
Answer: x[∞] = 2
Inverse Z-Transform
Partial Fraction Expansion
X(z) = \sum_{k} \frac{A_k z}{z - p_k} \Rightarrow x[n] = \sum_k A_k (p_k)^n u[n]
| Symbol | Description | Unit |
|---|---|---|
| p_k | k-th pole of X(z) | — |
| A_k | Residue at k-th pole | — |
Worked example
Find x[n] for X(z) = z^2/((z-1)(z-0.5)), causal system.
Given: Poles at z=1 and z=0.5
- Write X(z)/z = z/((z-1)(z-0.5))
- = A/(z-1) + B/(z-0.5)
- A = lim_{z->1} (z-1) * X(z)/z = 1/(1-0.5) = 2
- B = lim_{z->0.5} (z-0.5) * X(z)/z = 0.5/(0.5-1) = -1
- X(z) = 2z/(z-1) - z/(z-0.5)
- x[n] = [2(1)^n - (0.5)^n]u[n] = [2 - (0.5)^n]u[n]
Answer: x[n] = [2 - (0.5)^n]u[n]
Power Series (Long Division)
X(z) = x[0] + x[1]z^{-1} + x[2]z^{-2} + \cdots
| Symbol | Description | Unit |
|---|---|---|
| x[n] | Coefficient of z^{-n} in expansion | — |
Worked example
Find the first 3 samples of x[n] for X(z) = z/(z-0.5) using long division.
Given: X(z) = z/(z-0.5) = 1/(1-0.5z^{-1})
- Divide 1 by (1 - 0.5z^{-1}):
- 1st term: 1
- Remainder: 1 - (1 - 0.5z^{-1}) = 0.5z^{-1}
- 2nd term: 0.5z^{-1}
- Remainder: 0.5z^{-1} - 0.5z^{-1}(1 - 0.5z^{-1}) = 0.25z^{-2}
- 3rd term: 0.25z^{-2}
- X(z) ≈ 1 + 0.5z^{-1} + 0.25z^{-2} + ...
Answer: x[0]=1, x[1]=0.5, x[2]=0.25 — matches (0.5)^n
Transfer Function and System Analysis
System Transfer Function
H(z) = \frac{Y(z)}{X(z)} = \frac{\sum_{k=0}^{M} b_k z^{-k}}{1 + \sum_{k=1}^{N} a_k z^{-k}}
| Symbol | Description | Unit |
|---|---|---|
| b_k | Numerator (FIR) coefficients | — |
| a_k | Denominator (IIR) coefficients | — |
| H(z) | Discrete-time transfer function | — |
Worked example
A system has difference equation y[n] - 0.5y[n-1] = x[n] + 0.25x[n-1]. Find H(z).
Given: a1 = -0.5, b0 = 1, b1 = 0.25
- Take Z-transform both sides:
- Y(z) - 0.5z^{-1}Y(z) = X(z) + 0.25z^{-1}X(z)
- Y(z)(1 - 0.5z^{-1}) = X(z)(1 + 0.25z^{-1})
- H(z) = Y(z)/X(z) = (1 + 0.25z^{-1})/(1 - 0.5z^{-1})
- Multiply by z/z: H(z) = (z + 0.25)/(z - 0.5)
Answer: H(z) = (z + 0.25)/(z - 0.5)
Stability Condition
\text{BIBO Stable} \iff \text{All poles inside unit circle: } |p_k| < 1
| Symbol | Description | Unit |
|---|---|---|
| p_k | k-th pole of H(z) | — |
Worked example
Check stability of H(z) = (z+1)/((z-0.8)(z+1.2)).
Given: Poles at z = 0.8 and z = -1.2
- Check magnitudes of each pole
- |0.8| = 0.8 < 1 → this pole is inside unit circle (stable contribution)
- |-1.2| = 1.2 > 1 → this pole is outside unit circle
- Since at least one pole is outside unit circle, system is UNSTABLE
Answer: System is BIBO unstable due to pole at z = -1.2
Frequency Response from H(z)
H(e^{j\omega}) = H(z)\Big|_{z = e^{j\omega}}
| Symbol | Description | Unit |
|---|---|---|
| ω | Digital frequency | rad/sample |
| H(e^{jω}) | Discrete-time Fourier transform of h[n] | — |
Worked example
Find |H(e^{jω})| at ω=0 for H(z) = (z+1)/(z-0.5).
Given: ω = 0, so z = e^{j0} = 1
- Substitute z = e^{j0} = 1
- H(e^{j0}) = (1+1)/(1-0.5) = 2/0.5 = 4
- |H(e^{j0})| = 4
Answer: |H(e^{j0})| = 4 (DC gain = 4)
Difference Equations
General N-th Order Difference Equation
\sum_{k=0}^{N} a_k y[n-k] = \sum_{k=0}^{M} b_k x[n-k]
| Symbol | Description | Unit |
|---|---|---|
| a_k | Output coefficients, a0=1 | — |
| b_k | Input coefficients | — |
Worked example
Solve y[n] - 0.6y[n-1] = x[n] with x[n]=δ[n] and zero initial conditions.
Given: a1 = -0.6, b0 = 1, x[n] = δ[n]
- Take Z-transform: Y(z)(1 - 0.6z^{-1}) = 1
- Y(z) = 1/(1 - 0.6z^{-1}) = z/(z - 0.6)
- Inverse Z-transform: y[n] = (0.6)^n u[n]
- Verify: y[0] = 1, y[1] = 0.6, y[2] = 0.36 ✓
Answer: y[n] = (0.6)^n u[n]
Causal Recursive Computation
y[n] = -\sum_{k=1}^{N} a_k y[n-k] + \sum_{k=0}^{M} b_k x[n-k]
| Symbol | Description | Unit |
|---|---|---|
| y[n-k] | Past output samples | — |
| x[n-k] | Present and past input samples | — |
Worked example
For y[n] = 0.5y[n-1] + x[n], compute y[0], y[1], y[2] with x[n]=u[n], y[-1]=0.
Given: x[0]=x[1]=x[2]=1, y[-1]=0
- y[0] = 0.5*y[-1] + x[0] = 0.5*0 + 1 = 1
- y[1] = 0.5*y[0] + x[1] = 0.5*1 + 1 = 1.5
- y[2] = 0.5*y[1] + x[2] = 0.5*1.5 + 1 = 1.75
Answer: y[0]=1, y[1]=1.5, y[2]=1.75
Unilateral Z-Transform for Initial Conditions
Unilateral Time-Shift Property
\mathcal{Z}_u\{x[n-1]\} = z^{-1}X(z) + x[-1]
| Symbol | Description | Unit |
|---|---|---|
| x[-1] | Initial condition at n=-1 | — |
Worked example
Solve y[n] - 0.5y[n-1] = u[n] with y[-1] = 2.
Given: a1 = -0.5, y[-1] = 2, X(z) = z/(z-1)
- Apply unilateral Z-transform with IC:
- Y(z) - 0.5[z^{-1}Y(z) + y[-1]] = z/(z-1)
- Y(z) - 0.5z^{-1}Y(z) - 0.5*2 = z/(z-1)
- Y(z)(1 - 0.5z^{-1}) = z/(z-1) + 1
- Y(z) = [z/(z-1) + 1]/(1 - 0.5z^{-1})
- = z(z-1+z-1)/((z-1)(z-0.5)) ... simplify to get partial fractions
Answer: Y(z) = [z/(z-1) + 1] * z/(z-0.5) — decompose using partial fractions
Quick reference
| Formula | Expression |
|---|---|
| Bilateral Z-Transform | X(z) = \sum_{n=-\infty}^{\infty} x[n]z^{-n} |
| Z{a^n u[n]} | z/(z-a), |z|>|a| |
| Z{u[n]} | z/(z-1), |z|>1 |
| Z{δ[n]} | 1, all z |
| Z{n·u[n]} | z/(z-1)^2 |
| Time Shift | Z{x[n-k]} = z^{-k}X(z) |
| Convolution | x1*x2 ↔ X1(z)·X2(z) |
| Differentiation | Z{n·x[n]} = -z·dX/dz |
| Initial Value Theorem | x[0] = lim_{z→∞} X(z) |
| Final Value Theorem | x[∞] = lim_{z→1}(z-1)X(z) |
| Stability (causal) | All poles inside unit circle |
| Frequency Response | H(e^{jω}) = H(z)|_{z=e^{jω}} |
| Z{cos(ω₀n)u[n]} | z(z-cosω₀)/(z²-2z·cosω₀+1) |
| ROC Causal | |z| > r_max (outer pole magnitude) |
| ROC Anti-causal | |z| < r_min (inner pole magnitude) |
Exam tips
- GATE consistently tests ROC determination — always state ROC explicitly and check whether it includes the unit circle to determine BIBO stability.
- In partial fraction expansion, always divide X(z) by z first to get X(z)/z, perform PFD, then multiply back by z before taking inverse Z-transform.
- Examiners check whether the Final Value Theorem is correctly applied — verify that (z-1)X(z) has all poles strictly inside the unit circle before using it.
- The unilateral Z-transform is specifically tested for solving difference equations with non-zero initial conditions — do not ignore the initial condition terms.
- When finding the frequency response, substitute z=e^{jω} directly; common errors occur when students forget to rationalize the complex denominator to find magnitude.